The Problem

I have been given the following Lagrangian of a series of $N$ one-dimensional coupled oscillators, with distance a. I have also been given the boundary conditions: $y_0=0=y_{N+1},$ but have been given little explanation on where this came from. $$L=\sum_{i=1}^{N}\frac{1}{2}m\dot{y_i}^2-\sum_{i=0}^N\frac{1}{2}T\left(\frac{y_{i+1}-y_i}{a}\right)^2,$$ I am having trouble understanding how this lagrangian has been formed, in particular; why there is an a on the denominator of the second squared term.

My attempt at understanding:

I understand that the Lagrangian is expressed: $$L=KE-U,$$ The kinetic energy contribution of each of the $N$ coupled oscillators is simply: $$KE=\frac{1}{2}m\dot{y_i}^2,$$ Where $y_i$ is the displacement of the $i$th oscillator. Assuming all of the oscillators have equal mass, and none are fixed we should have the total kinetic energy: $$KE=\sum_{i=0}^{N}\frac{1}{2}m\dot{y_i}^2,$$ Generally, the potential energy of a spring is: $$U=\frac{1}{2}kx^2,$$ Where x represents the change in displacement, thus in our case the potential 'contributed' by each term is simply: $$U=\frac{1}{2}k(y_{i+1}-y_i^2),$$ because the terms in the bracket represent the 'stretch' of each of the springs, i.e. displacement. I assume the person who gave me the original Lagrangian has given me a spring constant of $T$ rather than $k$, but surely this should return a lagrangian: $$L=\sum_{i=0}^{N}\frac{1}{2}m\dot{y_i}^2-\sum_{i=0}^N\frac{1}{2}T\left(y_{i+1}-y_i\right)^2,$$ This is clearly not the same as what I was given, the first summation index is out by one place, I can assume this means that we are considering the first oscillator to be fixed, and thus the first velocity term goes to zero leaving: $$L=\sum_{i=1}^{N}\frac{1}{2}m\dot{y_i}^2-\sum_{i=0}^N\frac{1}{2}T\left(y_{i+1}-y_i\right)^2,$$ Dividing by a would give the ratio of the change w.r.t the original length of the spring, so I don't understand why this would be done? If anyone can see something that I am missing/done wrong, please let me know!
  • $\begingroup$ Related : Eigenvalue equation for kinetic and potential energy. $\endgroup$
    – Frobenius
    Oct 10 '20 at 14:31
  • $\begingroup$ In the original Lagrangian, $a$ is probably a typical length in the problem, so that $T$ has the dimension of an energy. Simply $T$ is not the spring constant $k$, but rather $k=T/a^2$. $\endgroup$
    – Quillo
    Oct 10 '20 at 15:34
  • $\begingroup$ T has the dimension of $N\,m$ but this is unusual $\endgroup$
    – Eli
    Oct 10 '20 at 15:47
  • $\begingroup$ @George Dixon k has the unit $N/m$ thus $T:=\frac Nm\,m^2=N\,m$ $\endgroup$
    – Eli
    Oct 10 '20 at 15:54
  • $\begingroup$ My mistake, So this means that my initial expression is correct? and the one that I have been given has for some reason has these two $a^2$ terms? (that cancel)... It may make sense in the context of what is written later in the notes, as it considers the limit as $a\rightarrow 0$, so in the form with a on the denominator it resembles a derivative $\endgroup$ Oct 10 '20 at 16:42

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